For a positive integer N, the digit-sum of N is defined as the sum of N itself and its digits. When M is the digitsum of N, we call N a generator of M.
For example, the digit-sum of 245 is 256 (= 245 + 2 + 4 + 5). Therefore, 245 is a generator of 256.
Not surprisingly, some numbers do not have any generators and some numbers have more than one generator. For example, the generators of 216 are 198 and 207.
You are to write a program to find the smallest generator of the given integer.
Input
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case takes one line containing an integer N, 1 ≤ N ≤ 100,000.
Output
Your program is to write to standard output. Print exactly one line for each test case. The line is to contain a generator of N for each test case. If N has multiple generators, print the smallest. If N does not have any generators, print 0.
The following shows sample input and output for three test cases.
Sample Input
3 216 121 2005
Output for the Sample Input
198 0 1979
要求是根據輸入的數找到其最小生成元，我一開始的思路是枚舉，但量太多，想用判斷來縮小枚舉範圍。設輸入數num，當num=99999時的各個位數的最大和爲45，也就是說只要在[num-45,num]這個區間進行枚舉，就可以比較輕鬆地得到答案。以下用C寫的答案：
#include<stdio.h>
int main()
{int ans=0,turns,num,i;
scanf("%d",&turns);while(turns>0){
scanf("%d",&num);
for(i=num-45;i<num;i++)
{if((i+i%10000/1000+i%1000/100+i%100/10+i%10)==num)
{ans=i; break;}}
if(!ans)printf("No answers\n");
else printf("%d\n",ans);turns--;ans=0;
}
return 0;
}